Sequences and Series | Mathematics | JEE Main Previous Year Questions

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Solution

Q.
Let $S_{k} = \frac{1 + 2 + . . . + k}{k}$ and $\sum_{j = 1}^{n} S_{j}^{2} = \frac{n}{A} (B n^{2} + C n + D)$ , where $A, B, C, D \in N$ and A has least value. Then [2023]

1 A + B is divisible by D

2 A + B + C + D is divisible by 5

3 A + C + D is not divisible by B

4 A + B = 5(D - C)

Ans.
(1)

We have, $S_{k} = \frac{1 + 2 + \dots + k}{k}$

$S_{k} = \frac{k (k + 1)}{2 k} = \frac{k + 1}{2}$

Now, $\sum_{j = 1}^{n} {(S_{j})}^{2} = \sum_{j = 1}^{n} {(\frac{j + 1}{2})}^{2} = \frac{1}{4} \sum_{j = 1}^{n} {(j + 1)}^{2}$

$= \frac{1}{4} [2^{2} + 3^{2} + 4^{2} + \dots + {(n + 1)}^{2}]$

   $= \frac{1}{4} [1^{2} + 2^{2} + 3^{2} + 4^{2} + \dots + {(n + 1)}^{2} - 1^{2}]$

   $= \frac{1}{4} [\frac{(n + 1) (n + 2) (2 n + 3)}{6} - 1]$

   $= \frac{1}{4} (\frac{2 n^{3} + 9 n^{2} + 13 n}{6}) = \frac{n}{24} (2 n^{2} + 9 n + 13)$

$= \frac{n}{A} (B n^{2} + C n + D)$

$∴$ $A = 24, B = 2, C = 9, D = 13$

$So, A + B = 24 + 2 = 26$

$∴$    $A + B is divisible by D .$

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